Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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The three edge integrals of z2 around the triangle with vertices 0, 1, and i sum to zero

Example

For the positively oriented boundary of Δ[0,1,i],

∫∂Δ[0,1,i]z2 dz=0.

More precisely, the integrals along the directed edges 0→1, 1→i, and i→0 are respectively

13,−i−13,i3.

Facts & Assumptions

Given: The oriented triangle Δ[0,1,i] and the integrand f(z)=z2.

[L1]

The oriented triangle boundary follows the directed edges 0→1→i→0 (Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter).

[L3]

If F is holomorphic with continuous derivative on an open set containing a rectifiable contour from u to v, then ∫F′=F(v)−F(u) (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path).

[L4]

Goursat's theorem gives zero integral for a holomorphic function around every filled triangle contained in its domain (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L5]

Verification

technique · direct
1.1L2L3L5

The polynomial F(z)=z3/3 is entire with F′(z)=z2; this derivative is holomorphic by [L2] and continuous by [L5], so every hypothesis of [L3] holds on each edge.

2.1step 1.1L1L3algebra

In the orientation of [L1], the endpoint increments are F(1)−F(0)=1/3, F(i)−F(1)=(−i−1)/3, and F(0)−F(i)=i/3.

3.1step 2.1L2L4algebra∎

Their sum is zero, proving the displayed integral directly; [L4] gives the same value because z2 is entire.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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